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Proof of Frullani's theorem
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How can I prove the Theorem of Frullani? I did not even know all the hypothesis that $f$ must satisfy, but I think that this are
Let $,f:left[ 0,infty right) to mathbb R$ be a a continuously differentiable function such that $$
mathop lim limits_x to infty fleft( x right) = 0,
$$
and let $
a,b in left( 0,infty right)$.
Prove that $$
intlimits_0^infty fracfleft( ax right) - fleft( bx right)
xdx = fleft( 0 right)left[ ln fracb
a right]
$$
If you know a more general version please give it to me )= I can´t prove it.
calculus real-analysis integration improper-integrals
edited Apr 24 at 14:03
GNU Supporter
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asked Sep 4 '11 at 14:53
August
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up vote
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favorite
How can I prove the Theorem of Frullani? I did not even know all the hypothesis that $f$ must satisfy, but I think that this are
Let $,f:left[ 0,infty right) to mathbb R$ be a a continuously differentiable function such that $$
mathop lim limits_x to infty fleft( x right) = 0,
$$
and let $
a,b in left( 0,infty right)$.
Prove that $$
intlimits_0^infty fracfleft( ax right) - fleft( bx right)
xdx = fleft( 0 right)left[ ln fracb
a right]
$$
If you know a more general version please give it to me )= I can´t prove it.
calculus real-analysis integration improper-integrals
edited Apr 24 at 14:03
GNU Supporter
11.8k72143
asked Sep 4 '11 at 14:53
August
1,42811623
The answers below are all good. There is an AMS article (1990) at: ams.org/journals/proc/1990-109-01/S0002-9939-1990-1007485-4 and it's free//available!
– rrogers
Apr 21 at 14:27
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up vote
93
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favorite
up vote
93
down vote
favorite
How can I prove the Theorem of Frullani? I did not even know all the hypothesis that $f$ must satisfy, but I think that this are
Let $,f:left[ 0,infty right) to mathbb R$ be a a continuously differentiable function such that $$
mathop lim limits_x to infty fleft( x right) = 0,
$$
and let $
a,b in left( 0,infty right)$.
Prove that $$
intlimits_0^infty fracfleft( ax right) - fleft( bx right)
xdx = fleft( 0 right)left[ ln fracb
a right]
$$
If you know a more general version please give it to me )= I can´t prove it.
calculus real-analysis integration improper-integrals
edited Apr 24 at 14:03
GNU Supporter
11.8k72143
asked Sep 4 '11 at 14:53
August
1,42811623
How can I prove the Theorem of Frullani? I did not even know all the hypothesis that $f$ must satisfy, but I think that this are
Let $,f:left[ 0,infty right) to mathbb R$ be a a continuously differentiable function such that $$
mathop lim limits_x to infty fleft( x right) = 0,
$$
and let $
a,b in left( 0,infty right)$.
Prove that $$
intlimits_0^infty fracfleft( ax right) - fleft( bx right)
xdx = fleft( 0 right)left[ ln fracb
a right]
$$
If you know a more general version please give it to me )= I can´t prove it.
calculus real-analysis integration improper-integrals
edited Apr 24 at 14:03
GNU Supporter
11.8k72143
asked Sep 4 '11 at 14:53
August
1,42811623
edited Apr 24 at 14:03
GNU Supporter
11.8k72143
edited Apr 24 at 14:03
GNU Supporter
11.8k72143
edited Apr 24 at 14:03
GNU Supporter
11.8k72143
11.8k72143
asked Sep 4 '11 at 14:53
August
1,42811623
asked Sep 4 '11 at 14:53
August
1,42811623
asked Sep 4 '11 at 14:53
August
1,42811623
1,42811623
The answers below are all good. There is an AMS article (1990) at: ams.org/journals/proc/1990-109-01/S0002-9939-1990-1007485-4 and it's free//available!
– rrogers
Apr 21 at 14:27
add a comment |Â
The answers below are all good. There is an AMS article (1990) at: ams.org/journals/proc/1990-109-01/S0002-9939-1990-1007485-4 and it's free//available!
– rrogers
Apr 21 at 14:27
The answers below are all good. There is an AMS article (1990) at: ams.org/journals/proc/1990-109-01/S0002-9939-1990-1007485-4 and it's free//available!
– rrogers
Apr 21 at 14:27
The answers below are all good. There is an AMS article (1990) at: ams.org/journals/proc/1990-109-01/S0002-9939-1990-1007485-4 and it's free//available!
– rrogers
Apr 21 at 14:27
add a comment |Â
7 Answers
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We will assume $a<b$.
Let $x,y>0$. We have:
beginalign*
int_x^ydfracf(at)-f(bt)tdt&=int_x^ydfracf(at)tdt-
int_x^ydfracf(bt)tdt\
&=int_ax^aydfracf(u)frac uafracdua-
int_bx^bydfracf(u)frac ubfracdub\
&=int_ax^aydfracf(u)udu-int_bx^bydfracf(u)udu\
&=int_ax^bxdfracf(u)udu+int_bx^aydfracf(u)udu
-int_bx^aydfracf(u)udu-int_ay^bydfracf(u)udu\
&=int_ax^bxdfracf(u)udu-int_ay^bydfracf(u)udu.
endalign*
Since $displaystyleint_0^+inftydfracf(at)-f(bt)tdt=lim_yto +inftylim_xto 0
int_x^ydfracf(at)-f(bt)tdt$ if these limits exist, we only have to show that the
limits $displaystylelim_xto 0int_ax^bxdfracf(u)udu$ and $displaystylelim_yto +inftyint_ay^bydfracf(u)udu$ exists, by computing them.
For the first, we denote $displaystyle m(x):=min_tinleft[ax,bxright]f(t)$ and
$displaystyle M(x):=max_tinleft[ax,bxright]f(t)$. We have for $x>0$:
$$m(x)lnleft(dfrac baright)leq int_ax^bxdfracf(u)uduleq
M(x)lnleft(dfrac baright) $$ and we get $displaystylelim_xto 0,m(x)=lim_xto 0, M(x)=f(0)$ thanks to the continuity of $f$.
For the second, fix $varepsilon>0$. We can find $x_0$ such that if $ugeq x_0$ then
$|f(u)|leq varepsilon$.
For $ygeq fracx_0a$, we get $displaystyleleft|int_ay^byfracf(u)uduright|
leq varepsilonlnleft(dfrac baright) $.
We notice that we didn't need the differentiability of $f$.
Added later, thanks to Didier's remark: if $f$ has a limit $l$ at $+infty$, then $gcolon
xmapsto f(x)-l$ is still continuous and has a limit $0$ at $+infty$. Then
$$int_0^+inftydfracf(at)-f(tb)tdt =
int_0^+inftydfracg(at)-g(tb)tdt =g(0)lnleft(dfrac baright) =
left(f(0)-lright)lnleft(dfrac baright).$$
answered Sep 4 '11 at 16:18
Davide Giraudo
121k15147249
2
+1. I like your solution. But you should replace lim m = lim M = 0 by lim m = lim M = f(0), to deduce that this part converges to f(0)log(b/a). // An extension is to assume that f has a limit at +oo, say f(+oo). Then your proof shows that the same result holds, with the limit (f(0)+f(+oo))log(b/a).
– Did
Sep 4 '11 at 16:45
@Didier: yes, it's of course $f(0)$ and I have corrected it. Maybe I should add the extension.
– Davide Giraudo
Sep 4 '11 at 16:49
2
It is enough to assume that $f$ is continuous on $(0,infty)$ and $lim_xto 0+ f(x)=:f(0+)$ is finite (and of course, $f$ has a finite limit in $+infty$). You can replace $f(0)$ by $f(0+)$ in the answer of @Giraudo.
– vesszabo
Aug 3 '12 at 20:46
2
@Davide Giraudo: awesome (+1)
– user 23571113
Aug 5 '12 at 14:52
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The result is true under weaker assumptions than you state, but under your conditions, there is a cute proof using double integrals. (To be on the safe side, assume that $f$ is monotone, or at least that $f' in L^1$. This will guarantee that we can change the order of integration.)
Let $D = (x,y) in mathbbR^2 : x ge 0, a le y le b $, and compute the integral
$$iint_D -f'(xy),dx,dy$$
in two different ways.
Firstly
beginalign
iint_D -f'(xy),dx,dy &= int_a^b left( int_0^infty -f'(xy),dx right),dy \ &= int_a^b left[ frac-f(xy)yright]^infty_0,dy \
&= int_a^b fracf(0)y,dy = f(0)(ln b - ln a).
endalign
On the other hand,
beginalign
iint_D -f'(xy),dx,dy &= int_0^infty left( int_a^b -f'(xy),dy right),dx\
&= int_0^infty left[ frac-f(xy)x right]_a^b,dx \
&= int_0^infty fracf(ax)-f(bx)x,dx.
endalign
answered Dec 30 '12 at 13:57
mrf
36.7k54484
(+1), but why do you need $f$ to be monotone? It's enough for $g=f'$ to be continuous, which is what the OP wrote (continuously differentiable)
– Alex
Apr 17 '15 at 9:04
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There is a claim that is slightly more general.
Let $f$ be such that $int_a^b f$ exists for each $a,b>0$. Suppose that $$A=lim_xto 0^+xint_x^1 fracf(t)t^2dt\B=lim_xto+inftyfrac 1 xint_1^x f(t)dt$$ exist.
Then $$int_0^inftyfracf(ax)-f(bx)xdx=(B-A)log frac ab$$
PROOF Define $xg(x)=displaystyle int_1^x f(t)dt$. Since $g'(x)+dfracg(x)x=dfracf(x)x$ we have $$int_a^b fracf(x)xdx=g(b)-g(a)+int_a^bfracg(x)xdx$$
Thus for $T>0$
$$int_Ta^Tb fracf(x)xdx=g(Tb)-g(Ta)+int_Ta^Tbfracg(x)xdx$$
But
$$int_Ta^Tbfracg(x)xdx-Bint_a^b fracdxx=int_a^bfracg(Tx)-Bxdx$$
Thus $$lim_Tto+inftyint_Ta^Tbfracg(x)xdx=Blogfrac ba$$ so
$$lim_Tto+inftyint_Ta^Tbfracf(x)xdx=Blogfrac ba$$
It follows, since $$int_1^Tfracf(ax)-f(bx)xdx=int_bT^aTfracf(x)xdx+int_a^b fracf(x)xdx$$ (note $a,b$ are swapped) that $$int_1^infty fracf(ax)-f(bx)xdx=Blogfrac ab+int_a^b fracf(x)xdx$$
Let $varepsilon >0$, $hat f(x)=f(1/x)$. Then $$intlimits_varepsilon ^1 fracfleft( x right)xdx = intlimits_1^varepsilon ^ - 1 frachat fleft( x right)xdx $$ and $$xintlimits_x^1 fracfleft( t right)t^2dt = frac1x^ - 1intlimits_1^x^ - 1 hat fleft( t right)dt = gleft( x^ - 1 right)$$
So $hat f(t)$ is in the hypothesis of the preceding work. It follows that $$lim_Tto+inftyintlimits_1^T frachat fleft( xa^ - 1 right) - hat fleft( xb^ - 1 right)x dx = Alog frac ba + intlimits_a^ - 1^b^ - 1 frachat fleft( x right)xdx $$
and by a change of variables $xmapsto x^-1$ we get $$intlimits_0^1 fracfleft( ax right) - fleft( bx right)x dx = Alog frac ba - intlimits_a^b fracfleft( x right)xdx $$ and summing gives the desired $$intlimits_0^infty fracfleft( ax right) - fleft( bx right)x dx = left( B - A right)log frac ab$$
This is due to T.M. Apostol.
OBS By L'Hôpital, if the limits at $x=0^+$ and $x=+infty$ exist, they equal $A$ and $B$ respectively.
answered Jul 23 '13 at 1:32
Pedro Tamaroff♦
93.8k10143290
Sorry for the (absurdly) late comment, but $ g $ may not be differentiable, so your justification of $int_a^bf(x)/x,dx=g(b)-g(a)+int_a^bg(x)/x,dx $, I think, is not quite right. It can be made more accurate, though. We have $int_a^b g(x)/x,dx=int_a^b x,g(x),d(-1/x)$, the last integral being Riemann-Stieltjes. Now, by integration by parts, $int_a^b g(x)/x,dx=g(a)-g(b)+int_a^bx^-1 ,d(x,g(x))$. Now, invoke Theorem 7.26 (Apostol's Analysis) to get $ int_a^b x^-1,d(x,g(x))=int_a^b f(x)/x,dx $.
– user149844
Jun 24 '17 at 4:06
@user149844 True. If $f$ is continuous then we're good.
– Pedro Tamaroff♦
Jun 24 '17 at 4:09
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You might be interested in an approach to Frullani's theorem I came across online. It is proven for the Lebesgue integral and the Denjoy-Perron integral. We are looking to prove the integral
beginequation*
int^infty_0fracf(ax)-f(bx)xdx=Aln(fracab)
endequation*
where $A$ is a constant. For the Lebesgue integral, the substitution $x=e^t,~alpha=ln(alpha),~beta=ln(b)$ is used to get
beginequation*
int^+infty_-infty f(e^t+alpha)-f(e^t+beta)dt=A(alpha-beta)
endequation*
which is equivalent to Frullani's theorem. Then verifying the integral
beginequation*
int^+infty_-infty g(x+alpha)-g(x+beta)dx=A(alpha-beta)
endequation*
for a Lebesgue integrable function $g:mathbbRtomathbbR~forall alpha,betain mathbbR$ will suffice. This is proved by setting an integrable function on the real line
beginequation*
h_alpha(x)=g(x+alpha)-g(x)~forallalphainmathbbR
endequation*
and applying the Fourier transform (as well as a little manipulation).
The Denjoy-Perron integral is used instead of the Lebesgue integral to avoid the problem of a locally integrable function $f:mathbbRtomathbbC$ admitting a derivative $f'(x)~forall xinmathbbR$ without $f'$ being locally integrable.
The case for the Denjoy-Perron integral is proved in a similar fashion.
Check out the following paper by J. Reyna
http://www.ams.org/journals/proc/1990-109-01/S0002-9939-1990-1007485-4/S0002-9939-1990-1007485-4.pdf
answered Jun 20 '15 at 20:32
George Simpson
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The following theorem is a beautiful generalization of Frullani’s integral theorem.
Let $f(x)-f(infty)=sum_k=0^inftyfracu(k)(-x)^kk!$ and $g(x)-g(infty)=sum_k=0^inftyfracv(k)(-x)^kk!$
$Theorem1$:
Let f, and g be continuous function on $[0,infty),$ assume that $f(0)=g(0)$ and $f(infty)=g(infty)$. Then if $a,b>0$
$$lim_n to 0+I_nequiv lim_n to 0+ int_0^inftyx^n-1lbrace f(ax)-g(bx) rbrace dx=lbrace f(0)-f(infty)rbrace bigg lbrace log bigg(fracba bigg)+fracddsbigg(logbigg(fracv(s)u(s)bigg) bigg)_s=0 bigg rbrace$$
if $f(x)=g(x),$ this theorem reduces to the Frullani’s theorem
$$int_0^infty fracf(ax)-f(bx)xdx=lbrace f(0)-f(infty) rbrace log bigg(fracba bigg).$$
Let prove $Theorem1$, To do this we need to use Ramanujan's master theorem , Which lies in the fact that
$$int_0^infty x^n-1sum_k=0^infty frac phi(k)(-x)^kk!dx= Gamma(n)phi(-n).$$
Applying the Master Theorem with $0<n<1,$ we find
$$I_n=int_0^infty x^n-1( f(ax)-g(bx))dx=int_0^infty x^n-1( lbrace f(ax)-f(infty) rbrace-lbrace g(bx)-g(infty) rbrace) dx$$
$$=Gamma(n)lbrace a^-nu(-n)-b^-nv(-n) rbrace$$
$$=Gamma(n+1) bigg lbrace fraca^-nu(-n)-b^-nv(-n)n bigg rbrace
$$
Letting $n$ tend to $0$, using L'Hospital's Rule and fact that $u(0)=v(0)=f(0)-f(infty).$ we deduce that
$$lim_n to inftyI_n=lim_n to infty bigg lbrace fracb^nv(n)-a^nu(n)n bigg rbrace$$
$$=lim_n to infty lbrace b^nv(n) log b+ b^nv'(n)-a^nu(n)log a-a^nu'(n)rbrace$$
$$= lbrace f(0)-f(infty) rbrace log bigg(fracba bigg)+v'(0)-u'(0)$$
$$=lbrace f(0)-f(infty) rbrace bigg lbrace log bigg(fracba bigg)+fracddsbigg(logbigg(fracv(s)u(s)bigg) bigg)_s=0 bigg rbrace$$
answered Mar 16 '16 at 17:37
vito
945815
2
This is incredible! May I ask is there a source for this? I'd like to know about some related analysis.
– Lee David Chung Lin
Mar 9 at 11:29
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On the assumption that $f$ is differentiable and $a,b>0$, an application of Fubini's theorem gives the result.
$$int_(0,infty) fracf(ax)-f(bx)x dx$$
$$=int_(0,infty) int_[bx,ax] f'(y) frac1x dy dx$$
Let $0<bx leq y leq ax$.
$$=int_(0,infty) int_frac1a y^frac1b y f'(y) frac1x dx dy$$
$$=int_(0,infty) f'(y) ln (fracab) dy$$
$$=(f(0)-f(infty)) ln fracba$$
answered Jul 23 '17 at 20:25
Ahmed S. Attaalla
13.8k11644
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The following is just a speeded-up version of the answer by Davide Giraudo.
Let $f(x)$ be a real-valued function defined for $xgeq 0$.
Suppose that $f(x)$ is Riemann integrable on every bounded interval
of nonnegative real numbers,
that $f(x)$ is continuous at $x=0$,
and that the limit $f(infty):=lim_xtoinfty f(x)$ exists
(as a finite quantity).
If $a>0$ and $b>0$, then the integral
beginequation*
int_,0^,infty fracf(ax)-f(bx)x dx tag1
endequation*
exists and is equal to $bigl(f(infty)-f(0)bigr),ln(a/b)$.
The assertion that the integral $(1)$ exists means
that the following limit exists,
beginequation*
lim_lto0,,htoinfty int_,l^,h fracf(ax)-f(bx)x dx
endequation*
where $l$ approaches $0$ independently of $h$ approaching $infty$,
and that this limit is the integral $(1)$.
Proof.$,$ Assume that $ageq b,$; this does not lose us any generality.
Let $0<l<h,$.
The change of variables $ax=by$ shows that
beginequation*
int_,l/a^,h/afracf(ax)xdx
~=~ int_,l/b^,h/bfracf(by)ydy~,
endequation*
so we have
beginalign*
int_,l/a^,h/afracf(ax)-f(bx)xdx
&~=~ int_,l/b^,h/bfracf(bx)xdx
~-~ int_,l/a^,h/afracf(bx)xdx \
&~=~ int_,h/a^,h/bfracf(bx)xdx
~-~ int_,l/a^,l/bfracf(bx)xdx~;
endalign*
we write the difference in the second line as $I(h) - I(l)$.
Let $varepsilon>0$.
There exists $l_varepsilon>0$ such that
$f(0)-varepsilonleq f(bx)leq f(0)+varepsilon$
for $0leq xleq l_varepsilon/b$.
Since by assumption $l/aleq l/b$ for every $l>0$, we obtain the estimate
beginequation*
int_,l/a^,l/bfracf(0)-varepsilonxdx
~leq~ I(l)
~leq~ int_,l/a^,l/bfracf(0)+varepsilonxdx
qquadqquad textfor $0<lleq l_varepsilon$~,
endequation*
that is,
beginequation*
bigl(f(0)-varepsilonbigr),lnfracab
~leq~ I(l)
~leq~ bigl(f(0)+varepsilonbigr),lnfracab
qquadqquad textfor $0<lleq l_varepsilon$~.
endequation*
In other words, $I(l)$ converges to $f(0),ln(a/b)$ as $l$ approaches $0$.
In the same way we see that $I(h)$ converges to $f(infty)$
as $h$ approaches $infty$.$,$ Done.
answered Jan 3 '17 at 14:57
chizhek
2,587622
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7 Answers
7
active
oldest
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7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
64
down vote
accepted
We will assume $a<b$.
Let $x,y>0$. We have:
beginalign*
int_x^ydfracf(at)-f(bt)tdt&=int_x^ydfracf(at)tdt-
int_x^ydfracf(bt)tdt\
&=int_ax^aydfracf(u)frac uafracdua-
int_bx^bydfracf(u)frac ubfracdub\
&=int_ax^aydfracf(u)udu-int_bx^bydfracf(u)udu\
&=int_ax^bxdfracf(u)udu+int_bx^aydfracf(u)udu
-int_bx^aydfracf(u)udu-int_ay^bydfracf(u)udu\
&=int_ax^bxdfracf(u)udu-int_ay^bydfracf(u)udu.
endalign*
Since $displaystyleint_0^+inftydfracf(at)-f(bt)tdt=lim_yto +inftylim_xto 0
int_x^ydfracf(at)-f(bt)tdt$ if these limits exist, we only have to show that the
limits $displaystylelim_xto 0int_ax^bxdfracf(u)udu$ and $displaystylelim_yto +inftyint_ay^bydfracf(u)udu$ exists, by computing them.
For the first, we denote $displaystyle m(x):=min_tinleft[ax,bxright]f(t)$ and
$displaystyle M(x):=max_tinleft[ax,bxright]f(t)$. We have for $x>0$:
$$m(x)lnleft(dfrac baright)leq int_ax^bxdfracf(u)uduleq
M(x)lnleft(dfrac baright) $$ and we get $displaystylelim_xto 0,m(x)=lim_xto 0, M(x)=f(0)$ thanks to the continuity of $f$.
For the second, fix $varepsilon>0$. We can find $x_0$ such that if $ugeq x_0$ then
$|f(u)|leq varepsilon$.
For $ygeq fracx_0a$, we get $displaystyleleft|int_ay^byfracf(u)uduright|
leq varepsilonlnleft(dfrac baright) $.
We notice that we didn't need the differentiability of $f$.
Added later, thanks to Didier's remark: if $f$ has a limit $l$ at $+infty$, then $gcolon
xmapsto f(x)-l$ is still continuous and has a limit $0$ at $+infty$. Then
$$int_0^+inftydfracf(at)-f(tb)tdt =
int_0^+inftydfracg(at)-g(tb)tdt =g(0)lnleft(dfrac baright) =
left(f(0)-lright)lnleft(dfrac baright).$$
answered Sep 4 '11 at 16:18
Davide Giraudo
121k15147249
2
+1. I like your solution. But you should replace lim m = lim M = 0 by lim m = lim M = f(0), to deduce that this part converges to f(0)log(b/a). // An extension is to assume that f has a limit at +oo, say f(+oo). Then your proof shows that the same result holds, with the limit (f(0)+f(+oo))log(b/a).
– Did
Sep 4 '11 at 16:45
@Didier: yes, it's of course $f(0)$ and I have corrected it. Maybe I should add the extension.
– Davide Giraudo
Sep 4 '11 at 16:49
2
It is enough to assume that $f$ is continuous on $(0,infty)$ and $lim_xto 0+ f(x)=:f(0+)$ is finite (and of course, $f$ has a finite limit in $+infty$). You can replace $f(0)$ by $f(0+)$ in the answer of @Giraudo.
– vesszabo
Aug 3 '12 at 20:46
2
@Davide Giraudo: awesome (+1)
– user 23571113
Aug 5 '12 at 14:52
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up vote
64
down vote
accepted
We will assume $a<b$.
Let $x,y>0$. We have:
beginalign*
int_x^ydfracf(at)-f(bt)tdt&=int_x^ydfracf(at)tdt-
int_x^ydfracf(bt)tdt\
&=int_ax^aydfracf(u)frac uafracdua-
int_bx^bydfracf(u)frac ubfracdub\
&=int_ax^aydfracf(u)udu-int_bx^bydfracf(u)udu\
&=int_ax^bxdfracf(u)udu+int_bx^aydfracf(u)udu
-int_bx^aydfracf(u)udu-int_ay^bydfracf(u)udu\
&=int_ax^bxdfracf(u)udu-int_ay^bydfracf(u)udu.
endalign*
Since $displaystyleint_0^+inftydfracf(at)-f(bt)tdt=lim_yto +inftylim_xto 0
int_x^ydfracf(at)-f(bt)tdt$ if these limits exist, we only have to show that the
limits $displaystylelim_xto 0int_ax^bxdfracf(u)udu$ and $displaystylelim_yto +inftyint_ay^bydfracf(u)udu$ exists, by computing them.
For the first, we denote $displaystyle m(x):=min_tinleft[ax,bxright]f(t)$ and
$displaystyle M(x):=max_tinleft[ax,bxright]f(t)$. We have for $x>0$:
$$m(x)lnleft(dfrac baright)leq int_ax^bxdfracf(u)uduleq
M(x)lnleft(dfrac baright) $$ and we get $displaystylelim_xto 0,m(x)=lim_xto 0, M(x)=f(0)$ thanks to the continuity of $f$.
For the second, fix $varepsilon>0$. We can find $x_0$ such that if $ugeq x_0$ then
$|f(u)|leq varepsilon$.
For $ygeq fracx_0a$, we get $displaystyleleft|int_ay^byfracf(u)uduright|
leq varepsilonlnleft(dfrac baright) $.
We notice that we didn't need the differentiability of $f$.
Added later, thanks to Didier's remark: if $f$ has a limit $l$ at $+infty$, then $gcolon
xmapsto f(x)-l$ is still continuous and has a limit $0$ at $+infty$. Then
$$int_0^+inftydfracf(at)-f(tb)tdt =
int_0^+inftydfracg(at)-g(tb)tdt =g(0)lnleft(dfrac baright) =
left(f(0)-lright)lnleft(dfrac baright).$$
answered Sep 4 '11 at 16:18
Davide Giraudo
121k15147249
2
+1. I like your solution. But you should replace lim m = lim M = 0 by lim m = lim M = f(0), to deduce that this part converges to f(0)log(b/a). // An extension is to assume that f has a limit at +oo, say f(+oo). Then your proof shows that the same result holds, with the limit (f(0)+f(+oo))log(b/a).
– Did
Sep 4 '11 at 16:45
@Didier: yes, it's of course $f(0)$ and I have corrected it. Maybe I should add the extension.
– Davide Giraudo
Sep 4 '11 at 16:49
2
It is enough to assume that $f$ is continuous on $(0,infty)$ and $lim_xto 0+ f(x)=:f(0+)$ is finite (and of course, $f$ has a finite limit in $+infty$). You can replace $f(0)$ by $f(0+)$ in the answer of @Giraudo.
– vesszabo
Aug 3 '12 at 20:46
2
@Davide Giraudo: awesome (+1)
– user 23571113
Aug 5 '12 at 14:52
add a comment |Â
up vote
64
down vote
accepted
up vote
64
down vote
accepted
We will assume $a<b$.
Let $x,y>0$. We have:
beginalign*
int_x^ydfracf(at)-f(bt)tdt&=int_x^ydfracf(at)tdt-
int_x^ydfracf(bt)tdt\
&=int_ax^aydfracf(u)frac uafracdua-
int_bx^bydfracf(u)frac ubfracdub\
&=int_ax^aydfracf(u)udu-int_bx^bydfracf(u)udu\
&=int_ax^bxdfracf(u)udu+int_bx^aydfracf(u)udu
-int_bx^aydfracf(u)udu-int_ay^bydfracf(u)udu\
&=int_ax^bxdfracf(u)udu-int_ay^bydfracf(u)udu.
endalign*
Since $displaystyleint_0^+inftydfracf(at)-f(bt)tdt=lim_yto +inftylim_xto 0
int_x^ydfracf(at)-f(bt)tdt$ if these limits exist, we only have to show that the
limits $displaystylelim_xto 0int_ax^bxdfracf(u)udu$ and $displaystylelim_yto +inftyint_ay^bydfracf(u)udu$ exists, by computing them.
For the first, we denote $displaystyle m(x):=min_tinleft[ax,bxright]f(t)$ and
$displaystyle M(x):=max_tinleft[ax,bxright]f(t)$. We have for $x>0$:
$$m(x)lnleft(dfrac baright)leq int_ax^bxdfracf(u)uduleq
M(x)lnleft(dfrac baright) $$ and we get $displaystylelim_xto 0,m(x)=lim_xto 0, M(x)=f(0)$ thanks to the continuity of $f$.
For the second, fix $varepsilon>0$. We can find $x_0$ such that if $ugeq x_0$ then
$|f(u)|leq varepsilon$.
For $ygeq fracx_0a$, we get $displaystyleleft|int_ay^byfracf(u)uduright|
leq varepsilonlnleft(dfrac baright) $.
We notice that we didn't need the differentiability of $f$.
Added later, thanks to Didier's remark: if $f$ has a limit $l$ at $+infty$, then $gcolon
xmapsto f(x)-l$ is still continuous and has a limit $0$ at $+infty$. Then
$$int_0^+inftydfracf(at)-f(tb)tdt =
int_0^+inftydfracg(at)-g(tb)tdt =g(0)lnleft(dfrac baright) =
left(f(0)-lright)lnleft(dfrac baright).$$
answered Sep 4 '11 at 16:18
Davide Giraudo
121k15147249
We will assume $a<b$.
Let $x,y>0$. We have:
beginalign*
int_x^ydfracf(at)-f(bt)tdt&=int_x^ydfracf(at)tdt-
int_x^ydfracf(bt)tdt\
&=int_ax^aydfracf(u)frac uafracdua-
int_bx^bydfracf(u)frac ubfracdub\
&=int_ax^aydfracf(u)udu-int_bx^bydfracf(u)udu\
&=int_ax^bxdfracf(u)udu+int_bx^aydfracf(u)udu
-int_bx^aydfracf(u)udu-int_ay^bydfracf(u)udu\
&=int_ax^bxdfracf(u)udu-int_ay^bydfracf(u)udu.
endalign*
Since $displaystyleint_0^+inftydfracf(at)-f(bt)tdt=lim_yto +inftylim_xto 0
int_x^ydfracf(at)-f(bt)tdt$ if these limits exist, we only have to show that the
limits $displaystylelim_xto 0int_ax^bxdfracf(u)udu$ and $displaystylelim_yto +inftyint_ay^bydfracf(u)udu$ exists, by computing them.
For the first, we denote $displaystyle m(x):=min_tinleft[ax,bxright]f(t)$ and
$displaystyle M(x):=max_tinleft[ax,bxright]f(t)$. We have for $x>0$:
$$m(x)lnleft(dfrac baright)leq int_ax^bxdfracf(u)uduleq
M(x)lnleft(dfrac baright) $$ and we get $displaystylelim_xto 0,m(x)=lim_xto 0, M(x)=f(0)$ thanks to the continuity of $f$.
For the second, fix $varepsilon>0$. We can find $x_0$ such that if $ugeq x_0$ then
$|f(u)|leq varepsilon$.
For $ygeq fracx_0a$, we get $displaystyleleft|int_ay^byfracf(u)uduright|
leq varepsilonlnleft(dfrac baright) $.
We notice that we didn't need the differentiability of $f$.
Added later, thanks to Didier's remark: if $f$ has a limit $l$ at $+infty$, then $gcolon
xmapsto f(x)-l$ is still continuous and has a limit $0$ at $+infty$. Then
$$int_0^+inftydfracf(at)-f(tb)tdt =
int_0^+inftydfracg(at)-g(tb)tdt =g(0)lnleft(dfrac baright) =
left(f(0)-lright)lnleft(dfrac baright).$$
answered Sep 4 '11 at 16:18
Davide Giraudo
121k15147249
edited Sep 4 '11 at 17:00
answered Sep 4 '11 at 16:18
Davide Giraudo
121k15147249
answered Sep 4 '11 at 16:18
Davide Giraudo
121k15147249
answered Sep 4 '11 at 16:18
Davide Giraudo
121k15147249
121k15147249
2
+1. I like your solution. But you should replace lim m = lim M = 0 by lim m = lim M = f(0), to deduce that this part converges to f(0)log(b/a). // An extension is to assume that f has a limit at +oo, say f(+oo). Then your proof shows that the same result holds, with the limit (f(0)+f(+oo))log(b/a).
– Did
Sep 4 '11 at 16:45
@Didier: yes, it's of course $f(0)$ and I have corrected it. Maybe I should add the extension.
– Davide Giraudo
Sep 4 '11 at 16:49
2
It is enough to assume that $f$ is continuous on $(0,infty)$ and $lim_xto 0+ f(x)=:f(0+)$ is finite (and of course, $f$ has a finite limit in $+infty$). You can replace $f(0)$ by $f(0+)$ in the answer of @Giraudo.
– vesszabo
Aug 3 '12 at 20:46
2
@Davide Giraudo: awesome (+1)
– user 23571113
Aug 5 '12 at 14:52
add a comment |Â
2
+1. I like your solution. But you should replace lim m = lim M = 0 by lim m = lim M = f(0), to deduce that this part converges to f(0)log(b/a). // An extension is to assume that f has a limit at +oo, say f(+oo). Then your proof shows that the same result holds, with the limit (f(0)+f(+oo))log(b/a).
– Did
Sep 4 '11 at 16:45
@Didier: yes, it's of course $f(0)$ and I have corrected it. Maybe I should add the extension.
– Davide Giraudo
Sep 4 '11 at 16:49
2
It is enough to assume that $f$ is continuous on $(0,infty)$ and $lim_xto 0+ f(x)=:f(0+)$ is finite (and of course, $f$ has a finite limit in $+infty$). You can replace $f(0)$ by $f(0+)$ in the answer of @Giraudo.
– vesszabo
Aug 3 '12 at 20:46
2
@Davide Giraudo: awesome (+1)
– user 23571113
Aug 5 '12 at 14:52
2
2
+1. I like your solution. But you should replace lim m = lim M = 0 by lim m = lim M = f(0), to deduce that this part converges to f(0)log(b/a). // An extension is to assume that f has a limit at +oo, say f(+oo). Then your proof shows that the same result holds, with the limit (f(0)+f(+oo))log(b/a).
– Did
Sep 4 '11 at 16:45
+1. I like your solution. But you should replace lim m = lim M = 0 by lim m = lim M = f(0), to deduce that this part converges to f(0)log(b/a). // An extension is to assume that f has a limit at +oo, say f(+oo). Then your proof shows that the same result holds, with the limit (f(0)+f(+oo))log(b/a).
– Did
Sep 4 '11 at 16:45
@Didier: yes, it's of course $f(0)$ and I have corrected it. Maybe I should add the extension.
– Davide Giraudo
Sep 4 '11 at 16:49
@Didier: yes, it's of course $f(0)$ and I have corrected it. Maybe I should add the extension.
– Davide Giraudo
Sep 4 '11 at 16:49
2
2
It is enough to assume that $f$ is continuous on $(0,infty)$ and $lim_xto 0+ f(x)=:f(0+)$ is finite (and of course, $f$ has a finite limit in $+infty$). You can replace $f(0)$ by $f(0+)$ in the answer of @Giraudo.
– vesszabo
Aug 3 '12 at 20:46
It is enough to assume that $f$ is continuous on $(0,infty)$ and $lim_xto 0+ f(x)=:f(0+)$ is finite (and of course, $f$ has a finite limit in $+infty$). You can replace $f(0)$ by $f(0+)$ in the answer of @Giraudo.
– vesszabo
Aug 3 '12 at 20:46
2
2
@Davide Giraudo: awesome (+1)
– user 23571113
Aug 5 '12 at 14:52
@Davide Giraudo: awesome (+1)
– user 23571113
Aug 5 '12 at 14:52
add a comment |Â
up vote
32
down vote
The result is true under weaker assumptions than you state, but under your conditions, there is a cute proof using double integrals. (To be on the safe side, assume that $f$ is monotone, or at least that $f' in L^1$. This will guarantee that we can change the order of integration.)
Let $D = (x,y) in mathbbR^2 : x ge 0, a le y le b $, and compute the integral
$$iint_D -f'(xy),dx,dy$$
in two different ways.
Firstly
beginalign
iint_D -f'(xy),dx,dy &= int_a^b left( int_0^infty -f'(xy),dx right),dy \ &= int_a^b left[ frac-f(xy)yright]^infty_0,dy \
&= int_a^b fracf(0)y,dy = f(0)(ln b - ln a).
endalign
On the other hand,
beginalign
iint_D -f'(xy),dx,dy &= int_0^infty left( int_a^b -f'(xy),dy right),dx\
&= int_0^infty left[ frac-f(xy)x right]_a^b,dx \
&= int_0^infty fracf(ax)-f(bx)x,dx.
endalign
answered Dec 30 '12 at 13:57
mrf
36.7k54484
(+1), but why do you need $f$ to be monotone? It's enough for $g=f'$ to be continuous, which is what the OP wrote (continuously differentiable)
– Alex
Apr 17 '15 at 9:04
add a comment |Â
up vote
32
down vote
The result is true under weaker assumptions than you state, but under your conditions, there is a cute proof using double integrals. (To be on the safe side, assume that $f$ is monotone, or at least that $f' in L^1$. This will guarantee that we can change the order of integration.)
Let $D = (x,y) in mathbbR^2 : x ge 0, a le y le b $, and compute the integral
$$iint_D -f'(xy),dx,dy$$
in two different ways.
Firstly
beginalign
iint_D -f'(xy),dx,dy &= int_a^b left( int_0^infty -f'(xy),dx right),dy \ &= int_a^b left[ frac-f(xy)yright]^infty_0,dy \
&= int_a^b fracf(0)y,dy = f(0)(ln b - ln a).
endalign
On the other hand,
beginalign
iint_D -f'(xy),dx,dy &= int_0^infty left( int_a^b -f'(xy),dy right),dx\
&= int_0^infty left[ frac-f(xy)x right]_a^b,dx \
&= int_0^infty fracf(ax)-f(bx)x,dx.
endalign
answered Dec 30 '12 at 13:57
mrf
36.7k54484
(+1), but why do you need $f$ to be monotone? It's enough for $g=f'$ to be continuous, which is what the OP wrote (continuously differentiable)
– Alex
Apr 17 '15 at 9:04
add a comment |Â
up vote
32
down vote
up vote
32
down vote
The result is true under weaker assumptions than you state, but under your conditions, there is a cute proof using double integrals. (To be on the safe side, assume that $f$ is monotone, or at least that $f' in L^1$. This will guarantee that we can change the order of integration.)
Let $D = (x,y) in mathbbR^2 : x ge 0, a le y le b $, and compute the integral
$$iint_D -f'(xy),dx,dy$$
in two different ways.
Firstly
beginalign
iint_D -f'(xy),dx,dy &= int_a^b left( int_0^infty -f'(xy),dx right),dy \ &= int_a^b left[ frac-f(xy)yright]^infty_0,dy \
&= int_a^b fracf(0)y,dy = f(0)(ln b - ln a).
endalign
On the other hand,
beginalign
iint_D -f'(xy),dx,dy &= int_0^infty left( int_a^b -f'(xy),dy right),dx\
&= int_0^infty left[ frac-f(xy)x right]_a^b,dx \
&= int_0^infty fracf(ax)-f(bx)x,dx.
endalign
answered Dec 30 '12 at 13:57
mrf
36.7k54484
The result is true under weaker assumptions than you state, but under your conditions, there is a cute proof using double integrals. (To be on the safe side, assume that $f$ is monotone, or at least that $f' in L^1$. This will guarantee that we can change the order of integration.)
Let $D = (x,y) in mathbbR^2 : x ge 0, a le y le b $, and compute the integral
$$iint_D -f'(xy),dx,dy$$
in two different ways.
Firstly
beginalign
iint_D -f'(xy),dx,dy &= int_a^b left( int_0^infty -f'(xy),dx right),dy \ &= int_a^b left[ frac-f(xy)yright]^infty_0,dy \
&= int_a^b fracf(0)y,dy = f(0)(ln b - ln a).
endalign
On the other hand,
beginalign
iint_D -f'(xy),dx,dy &= int_0^infty left( int_a^b -f'(xy),dy right),dx\
&= int_0^infty left[ frac-f(xy)x right]_a^b,dx \
&= int_0^infty fracf(ax)-f(bx)x,dx.
endalign
answered Dec 30 '12 at 13:57
mrf
36.7k54484
edited Jan 16 '13 at 22:46
answered Dec 30 '12 at 13:57
mrf
36.7k54484
answered Dec 30 '12 at 13:57
mrf
36.7k54484
answered Dec 30 '12 at 13:57
mrf
36.7k54484
36.7k54484
(+1), but why do you need $f$ to be monotone? It's enough for $g=f'$ to be continuous, which is what the OP wrote (continuously differentiable)
– Alex
Apr 17 '15 at 9:04
add a comment |Â
(+1), but why do you need $f$ to be monotone? It's enough for $g=f'$ to be continuous, which is what the OP wrote (continuously differentiable)
– Alex
Apr 17 '15 at 9:04
(+1), but why do you need $f$ to be monotone? It's enough for $g=f'$ to be continuous, which is what the OP wrote (continuously differentiable)
– Alex
Apr 17 '15 at 9:04
(+1), but why do you need $f$ to be monotone? It's enough for $g=f'$ to be continuous, which is what the OP wrote (continuously differentiable)
– Alex
Apr 17 '15 at 9:04
add a comment |Â
up vote
27
down vote
There is a claim that is slightly more general.
Let $f$ be such that $int_a^b f$ exists for each $a,b>0$. Suppose that $$A=lim_xto 0^+xint_x^1 fracf(t)t^2dt\B=lim_xto+inftyfrac 1 xint_1^x f(t)dt$$ exist.
Then $$int_0^inftyfracf(ax)-f(bx)xdx=(B-A)log frac ab$$
PROOF Define $xg(x)=displaystyle int_1^x f(t)dt$. Since $g'(x)+dfracg(x)x=dfracf(x)x$ we have $$int_a^b fracf(x)xdx=g(b)-g(a)+int_a^bfracg(x)xdx$$
Thus for $T>0$
$$int_Ta^Tb fracf(x)xdx=g(Tb)-g(Ta)+int_Ta^Tbfracg(x)xdx$$
But
$$int_Ta^Tbfracg(x)xdx-Bint_a^b fracdxx=int_a^bfracg(Tx)-Bxdx$$
Thus $$lim_Tto+inftyint_Ta^Tbfracg(x)xdx=Blogfrac ba$$ so
$$lim_Tto+inftyint_Ta^Tbfracf(x)xdx=Blogfrac ba$$
It follows, since $$int_1^Tfracf(ax)-f(bx)xdx=int_bT^aTfracf(x)xdx+int_a^b fracf(x)xdx$$ (note $a,b$ are swapped) that $$int_1^infty fracf(ax)-f(bx)xdx=Blogfrac ab+int_a^b fracf(x)xdx$$
Let $varepsilon >0$, $hat f(x)=f(1/x)$. Then $$intlimits_varepsilon ^1 fracfleft( x right)xdx = intlimits_1^varepsilon ^ - 1 frachat fleft( x right)xdx $$ and $$xintlimits_x^1 fracfleft( t right)t^2dt = frac1x^ - 1intlimits_1^x^ - 1 hat fleft( t right)dt = gleft( x^ - 1 right)$$
So $hat f(t)$ is in the hypothesis of the preceding work. It follows that $$lim_Tto+inftyintlimits_1^T frachat fleft( xa^ - 1 right) - hat fleft( xb^ - 1 right)x dx = Alog frac ba + intlimits_a^ - 1^b^ - 1 frachat fleft( x right)xdx $$
and by a change of variables $xmapsto x^-1$ we get $$intlimits_0^1 fracfleft( ax right) - fleft( bx right)x dx = Alog frac ba - intlimits_a^b fracfleft( x right)xdx $$ and summing gives the desired $$intlimits_0^infty fracfleft( ax right) - fleft( bx right)x dx = left( B - A right)log frac ab$$
This is due to T.M. Apostol.
OBS By L'Hôpital, if the limits at $x=0^+$ and $x=+infty$ exist, they equal $A$ and $B$ respectively.
answered Jul 23 '13 at 1:32
Pedro Tamaroff♦
93.8k10143290
Sorry for the (absurdly) late comment, but $ g $ may not be differentiable, so your justification of $int_a^bf(x)/x,dx=g(b)-g(a)+int_a^bg(x)/x,dx $, I think, is not quite right. It can be made more accurate, though. We have $int_a^b g(x)/x,dx=int_a^b x,g(x),d(-1/x)$, the last integral being Riemann-Stieltjes. Now, by integration by parts, $int_a^b g(x)/x,dx=g(a)-g(b)+int_a^bx^-1 ,d(x,g(x))$. Now, invoke Theorem 7.26 (Apostol's Analysis) to get $ int_a^b x^-1,d(x,g(x))=int_a^b f(x)/x,dx $.
– user149844
Jun 24 '17 at 4:06
@user149844 True. If $f$ is continuous then we're good.
– Pedro Tamaroff♦
Jun 24 '17 at 4:09
add a comment |Â
up vote
27
down vote
There is a claim that is slightly more general.
Let $f$ be such that $int_a^b f$ exists for each $a,b>0$. Suppose that $$A=lim_xto 0^+xint_x^1 fracf(t)t^2dt\B=lim_xto+inftyfrac 1 xint_1^x f(t)dt$$ exist.
Then $$int_0^inftyfracf(ax)-f(bx)xdx=(B-A)log frac ab$$
PROOF Define $xg(x)=displaystyle int_1^x f(t)dt$. Since $g'(x)+dfracg(x)x=dfracf(x)x$ we have $$int_a^b fracf(x)xdx=g(b)-g(a)+int_a^bfracg(x)xdx$$
Thus for $T>0$
$$int_Ta^Tb fracf(x)xdx=g(Tb)-g(Ta)+int_Ta^Tbfracg(x)xdx$$
But
$$int_Ta^Tbfracg(x)xdx-Bint_a^b fracdxx=int_a^bfracg(Tx)-Bxdx$$
Thus $$lim_Tto+inftyint_Ta^Tbfracg(x)xdx=Blogfrac ba$$ so
$$lim_Tto+inftyint_Ta^Tbfracf(x)xdx=Blogfrac ba$$
It follows, since $$int_1^Tfracf(ax)-f(bx)xdx=int_bT^aTfracf(x)xdx+int_a^b fracf(x)xdx$$ (note $a,b$ are swapped) that $$int_1^infty fracf(ax)-f(bx)xdx=Blogfrac ab+int_a^b fracf(x)xdx$$
Let $varepsilon >0$, $hat f(x)=f(1/x)$. Then $$intlimits_varepsilon ^1 fracfleft( x right)xdx = intlimits_1^varepsilon ^ - 1 frachat fleft( x right)xdx $$ and $$xintlimits_x^1 fracfleft( t right)t^2dt = frac1x^ - 1intlimits_1^x^ - 1 hat fleft( t right)dt = gleft( x^ - 1 right)$$
So $hat f(t)$ is in the hypothesis of the preceding work. It follows that $$lim_Tto+inftyintlimits_1^T frachat fleft( xa^ - 1 right) - hat fleft( xb^ - 1 right)x dx = Alog frac ba + intlimits_a^ - 1^b^ - 1 frachat fleft( x right)xdx $$
and by a change of variables $xmapsto x^-1$ we get $$intlimits_0^1 fracfleft( ax right) - fleft( bx right)x dx = Alog frac ba - intlimits_a^b fracfleft( x right)xdx $$ and summing gives the desired $$intlimits_0^infty fracfleft( ax right) - fleft( bx right)x dx = left( B - A right)log frac ab$$
This is due to T.M. Apostol.
OBS By L'Hôpital, if the limits at $x=0^+$ and $x=+infty$ exist, they equal $A$ and $B$ respectively.
answered Jul 23 '13 at 1:32
Pedro Tamaroff♦
93.8k10143290
Sorry for the (absurdly) late comment, but $ g $ may not be differentiable, so your justification of $int_a^bf(x)/x,dx=g(b)-g(a)+int_a^bg(x)/x,dx $, I think, is not quite right. It can be made more accurate, though. We have $int_a^b g(x)/x,dx=int_a^b x,g(x),d(-1/x)$, the last integral being Riemann-Stieltjes. Now, by integration by parts, $int_a^b g(x)/x,dx=g(a)-g(b)+int_a^bx^-1 ,d(x,g(x))$. Now, invoke Theorem 7.26 (Apostol's Analysis) to get $ int_a^b x^-1,d(x,g(x))=int_a^b f(x)/x,dx $.
– user149844
Jun 24 '17 at 4:06
@user149844 True. If $f$ is continuous then we're good.
– Pedro Tamaroff♦
Jun 24 '17 at 4:09
add a comment |Â
up vote
27
down vote
up vote
27
down vote
There is a claim that is slightly more general.
Let $f$ be such that $int_a^b f$ exists for each $a,b>0$. Suppose that $$A=lim_xto 0^+xint_x^1 fracf(t)t^2dt\B=lim_xto+inftyfrac 1 xint_1^x f(t)dt$$ exist.
Then $$int_0^inftyfracf(ax)-f(bx)xdx=(B-A)log frac ab$$
PROOF Define $xg(x)=displaystyle int_1^x f(t)dt$. Since $g'(x)+dfracg(x)x=dfracf(x)x$ we have $$int_a^b fracf(x)xdx=g(b)-g(a)+int_a^bfracg(x)xdx$$
Thus for $T>0$
$$int_Ta^Tb fracf(x)xdx=g(Tb)-g(Ta)+int_Ta^Tbfracg(x)xdx$$
But
$$int_Ta^Tbfracg(x)xdx-Bint_a^b fracdxx=int_a^bfracg(Tx)-Bxdx$$
Thus $$lim_Tto+inftyint_Ta^Tbfracg(x)xdx=Blogfrac ba$$ so
$$lim_Tto+inftyint_Ta^Tbfracf(x)xdx=Blogfrac ba$$
It follows, since $$int_1^Tfracf(ax)-f(bx)xdx=int_bT^aTfracf(x)xdx+int_a^b fracf(x)xdx$$ (note $a,b$ are swapped) that $$int_1^infty fracf(ax)-f(bx)xdx=Blogfrac ab+int_a^b fracf(x)xdx$$
Let $varepsilon >0$, $hat f(x)=f(1/x)$. Then $$intlimits_varepsilon ^1 fracfleft( x right)xdx = intlimits_1^varepsilon ^ - 1 frachat fleft( x right)xdx $$ and $$xintlimits_x^1 fracfleft( t right)t^2dt = frac1x^ - 1intlimits_1^x^ - 1 hat fleft( t right)dt = gleft( x^ - 1 right)$$
So $hat f(t)$ is in the hypothesis of the preceding work. It follows that $$lim_Tto+inftyintlimits_1^T frachat fleft( xa^ - 1 right) - hat fleft( xb^ - 1 right)x dx = Alog frac ba + intlimits_a^ - 1^b^ - 1 frachat fleft( x right)xdx $$
and by a change of variables $xmapsto x^-1$ we get $$intlimits_0^1 fracfleft( ax right) - fleft( bx right)x dx = Alog frac ba - intlimits_a^b fracfleft( x right)xdx $$ and summing gives the desired $$intlimits_0^infty fracfleft( ax right) - fleft( bx right)x dx = left( B - A right)log frac ab$$
This is due to T.M. Apostol.
OBS By L'Hôpital, if the limits at $x=0^+$ and $x=+infty$ exist, they equal $A$ and $B$ respectively.
answered Jul 23 '13 at 1:32
Pedro Tamaroff♦
93.8k10143290
There is a claim that is slightly more general.
Let $f$ be such that $int_a^b f$ exists for each $a,b>0$. Suppose that $$A=lim_xto 0^+xint_x^1 fracf(t)t^2dt\B=lim_xto+inftyfrac 1 xint_1^x f(t)dt$$ exist.
Then $$int_0^inftyfracf(ax)-f(bx)xdx=(B-A)log frac ab$$
PROOF Define $xg(x)=displaystyle int_1^x f(t)dt$. Since $g'(x)+dfracg(x)x=dfracf(x)x$ we have $$int_a^b fracf(x)xdx=g(b)-g(a)+int_a^bfracg(x)xdx$$
Thus for $T>0$
$$int_Ta^Tb fracf(x)xdx=g(Tb)-g(Ta)+int_Ta^Tbfracg(x)xdx$$
But
$$int_Ta^Tbfracg(x)xdx-Bint_a^b fracdxx=int_a^bfracg(Tx)-Bxdx$$
Thus $$lim_Tto+inftyint_Ta^Tbfracg(x)xdx=Blogfrac ba$$ so
$$lim_Tto+inftyint_Ta^Tbfracf(x)xdx=Blogfrac ba$$
It follows, since $$int_1^Tfracf(ax)-f(bx)xdx=int_bT^aTfracf(x)xdx+int_a^b fracf(x)xdx$$ (note $a,b$ are swapped) that $$int_1^infty fracf(ax)-f(bx)xdx=Blogfrac ab+int_a^b fracf(x)xdx$$
Let $varepsilon >0$, $hat f(x)=f(1/x)$. Then $$intlimits_varepsilon ^1 fracfleft( x right)xdx = intlimits_1^varepsilon ^ - 1 frachat fleft( x right)xdx $$ and $$xintlimits_x^1 fracfleft( t right)t^2dt = frac1x^ - 1intlimits_1^x^ - 1 hat fleft( t right)dt = gleft( x^ - 1 right)$$
So $hat f(t)$ is in the hypothesis of the preceding work. It follows that $$lim_Tto+inftyintlimits_1^T frachat fleft( xa^ - 1 right) - hat fleft( xb^ - 1 right)x dx = Alog frac ba + intlimits_a^ - 1^b^ - 1 frachat fleft( x right)xdx $$
and by a change of variables $xmapsto x^-1$ we get $$intlimits_0^1 fracfleft( ax right) - fleft( bx right)x dx = Alog frac ba - intlimits_a^b fracfleft( x right)xdx $$ and summing gives the desired $$intlimits_0^infty fracfleft( ax right) - fleft( bx right)x dx = left( B - A right)log frac ab$$
This is due to T.M. Apostol.
OBS By L'Hôpital, if the limits at $x=0^+$ and $x=+infty$ exist, they equal $A$ and $B$ respectively.
answered Jul 23 '13 at 1:32
Pedro Tamaroff♦
93.8k10143290
edited Jul 23 '13 at 1:47
answered Jul 23 '13 at 1:32
Pedro Tamaroff♦
93.8k10143290
answered Jul 23 '13 at 1:32
Pedro Tamaroff♦
93.8k10143290
answered Jul 23 '13 at 1:32
Pedro Tamaroff♦
93.8k10143290
93.8k10143290
Sorry for the (absurdly) late comment, but $ g $ may not be differentiable, so your justification of $int_a^bf(x)/x,dx=g(b)-g(a)+int_a^bg(x)/x,dx $, I think, is not quite right. It can be made more accurate, though. We have $int_a^b g(x)/x,dx=int_a^b x,g(x),d(-1/x)$, the last integral being Riemann-Stieltjes. Now, by integration by parts, $int_a^b g(x)/x,dx=g(a)-g(b)+int_a^bx^-1 ,d(x,g(x))$. Now, invoke Theorem 7.26 (Apostol's Analysis) to get $ int_a^b x^-1,d(x,g(x))=int_a^b f(x)/x,dx $.
– user149844
Jun 24 '17 at 4:06
@user149844 True. If $f$ is continuous then we're good.
– Pedro Tamaroff♦
Jun 24 '17 at 4:09
add a comment |Â
Sorry for the (absurdly) late comment, but $ g $ may not be differentiable, so your justification of $int_a^bf(x)/x,dx=g(b)-g(a)+int_a^bg(x)/x,dx $, I think, is not quite right. It can be made more accurate, though. We have $int_a^b g(x)/x,dx=int_a^b x,g(x),d(-1/x)$, the last integral being Riemann-Stieltjes. Now, by integration by parts, $int_a^b g(x)/x,dx=g(a)-g(b)+int_a^bx^-1 ,d(x,g(x))$. Now, invoke Theorem 7.26 (Apostol's Analysis) to get $ int_a^b x^-1,d(x,g(x))=int_a^b f(x)/x,dx $.
– user149844
Jun 24 '17 at 4:06
@user149844 True. If $f$ is continuous then we're good.
– Pedro Tamaroff♦
Jun 24 '17 at 4:09
Sorry for the (absurdly) late comment, but $ g $ may not be differentiable, so your justification of $int_a^bf(x)/x,dx=g(b)-g(a)+int_a^bg(x)/x,dx $, I think, is not quite right. It can be made more accurate, though. We have $int_a^b g(x)/x,dx=int_a^b x,g(x),d(-1/x)$, the last integral being Riemann-Stieltjes. Now, by integration by parts, $int_a^b g(x)/x,dx=g(a)-g(b)+int_a^bx^-1 ,d(x,g(x))$. Now, invoke Theorem 7.26 (Apostol's Analysis) to get $ int_a^b x^-1,d(x,g(x))=int_a^b f(x)/x,dx $.
– user149844
Jun 24 '17 at 4:06
Sorry for the (absurdly) late comment, but $ g $ may not be differentiable, so your justification of $int_a^bf(x)/x,dx=g(b)-g(a)+int_a^bg(x)/x,dx $, I think, is not quite right. It can be made more accurate, though. We have $int_a^b g(x)/x,dx=int_a^b x,g(x),d(-1/x)$, the last integral being Riemann-Stieltjes. Now, by integration by parts, $int_a^b g(x)/x,dx=g(a)-g(b)+int_a^bx^-1 ,d(x,g(x))$. Now, invoke Theorem 7.26 (Apostol's Analysis) to get $ int_a^b x^-1,d(x,g(x))=int_a^b f(x)/x,dx $.
– user149844
Jun 24 '17 at 4:06
@user149844 True. If $f$ is continuous then we're good.
– Pedro Tamaroff♦
Jun 24 '17 at 4:09
@user149844 True. If $f$ is continuous then we're good.
– Pedro Tamaroff♦
Jun 24 '17 at 4:09
add a comment |Â
up vote
9
down vote
You might be interested in an approach to Frullani's theorem I came across online. It is proven for the Lebesgue integral and the Denjoy-Perron integral. We are looking to prove the integral
beginequation*
int^infty_0fracf(ax)-f(bx)xdx=Aln(fracab)
endequation*
where $A$ is a constant. For the Lebesgue integral, the substitution $x=e^t,~alpha=ln(alpha),~beta=ln(b)$ is used to get
beginequation*
int^+infty_-infty f(e^t+alpha)-f(e^t+beta)dt=A(alpha-beta)
endequation*
which is equivalent to Frullani's theorem. Then verifying the integral
beginequation*
int^+infty_-infty g(x+alpha)-g(x+beta)dx=A(alpha-beta)
endequation*
for a Lebesgue integrable function $g:mathbbRtomathbbR~forall alpha,betain mathbbR$ will suffice. This is proved by setting an integrable function on the real line
beginequation*
h_alpha(x)=g(x+alpha)-g(x)~forallalphainmathbbR
endequation*
and applying the Fourier transform (as well as a little manipulation).
The Denjoy-Perron integral is used instead of the Lebesgue integral to avoid the problem of a locally integrable function $f:mathbbRtomathbbC$ admitting a derivative $f'(x)~forall xinmathbbR$ without $f'$ being locally integrable.
The case for the Denjoy-Perron integral is proved in a similar fashion.
Check out the following paper by J. Reyna
http://www.ams.org/journals/proc/1990-109-01/S0002-9939-1990-1007485-4/S0002-9939-1990-1007485-4.pdf
answered Jun 20 '15 at 20:32
George Simpson
4,56432449
add a comment |Â
up vote
9
down vote
You might be interested in an approach to Frullani's theorem I came across online. It is proven for the Lebesgue integral and the Denjoy-Perron integral. We are looking to prove the integral
beginequation*
int^infty_0fracf(ax)-f(bx)xdx=Aln(fracab)
endequation*
where $A$ is a constant. For the Lebesgue integral, the substitution $x=e^t,~alpha=ln(alpha),~beta=ln(b)$ is used to get
beginequation*
int^+infty_-infty f(e^t+alpha)-f(e^t+beta)dt=A(alpha-beta)
endequation*
which is equivalent to Frullani's theorem. Then verifying the integral
beginequation*
int^+infty_-infty g(x+alpha)-g(x+beta)dx=A(alpha-beta)
endequation*
for a Lebesgue integrable function $g:mathbbRtomathbbR~forall alpha,betain mathbbR$ will suffice. This is proved by setting an integrable function on the real line
beginequation*
h_alpha(x)=g(x+alpha)-g(x)~forallalphainmathbbR
endequation*
and applying the Fourier transform (as well as a little manipulation).
The Denjoy-Perron integral is used instead of the Lebesgue integral to avoid the problem of a locally integrable function $f:mathbbRtomathbbC$ admitting a derivative $f'(x)~forall xinmathbbR$ without $f'$ being locally integrable.
The case for the Denjoy-Perron integral is proved in a similar fashion.
Check out the following paper by J. Reyna
http://www.ams.org/journals/proc/1990-109-01/S0002-9939-1990-1007485-4/S0002-9939-1990-1007485-4.pdf
answered Jun 20 '15 at 20:32
George Simpson
4,56432449
add a comment |Â
up vote
9
down vote
up vote
9
down vote
You might be interested in an approach to Frullani's theorem I came across online. It is proven for the Lebesgue integral and the Denjoy-Perron integral. We are looking to prove the integral
beginequation*
int^infty_0fracf(ax)-f(bx)xdx=Aln(fracab)
endequation*
where $A$ is a constant. For the Lebesgue integral, the substitution $x=e^t,~alpha=ln(alpha),~beta=ln(b)$ is used to get
beginequation*
int^+infty_-infty f(e^t+alpha)-f(e^t+beta)dt=A(alpha-beta)
endequation*
which is equivalent to Frullani's theorem. Then verifying the integral
beginequation*
int^+infty_-infty g(x+alpha)-g(x+beta)dx=A(alpha-beta)
endequation*
for a Lebesgue integrable function $g:mathbbRtomathbbR~forall alpha,betain mathbbR$ will suffice. This is proved by setting an integrable function on the real line
beginequation*
h_alpha(x)=g(x+alpha)-g(x)~forallalphainmathbbR
endequation*
and applying the Fourier transform (as well as a little manipulation).
The Denjoy-Perron integral is used instead of the Lebesgue integral to avoid the problem of a locally integrable function $f:mathbbRtomathbbC$ admitting a derivative $f'(x)~forall xinmathbbR$ without $f'$ being locally integrable.
The case for the Denjoy-Perron integral is proved in a similar fashion.
Check out the following paper by J. Reyna
http://www.ams.org/journals/proc/1990-109-01/S0002-9939-1990-1007485-4/S0002-9939-1990-1007485-4.pdf
answered Jun 20 '15 at 20:32
George Simpson
4,56432449
You might be interested in an approach to Frullani's theorem I came across online. It is proven for the Lebesgue integral and the Denjoy-Perron integral. We are looking to prove the integral
beginequation*
int^infty_0fracf(ax)-f(bx)xdx=Aln(fracab)
endequation*
where $A$ is a constant. For the Lebesgue integral, the substitution $x=e^t,~alpha=ln(alpha),~beta=ln(b)$ is used to get
beginequation*
int^+infty_-infty f(e^t+alpha)-f(e^t+beta)dt=A(alpha-beta)
endequation*
which is equivalent to Frullani's theorem. Then verifying the integral
beginequation*
int^+infty_-infty g(x+alpha)-g(x+beta)dx=A(alpha-beta)
endequation*
for a Lebesgue integrable function $g:mathbbRtomathbbR~forall alpha,betain mathbbR$ will suffice. This is proved by setting an integrable function on the real line
beginequation*
h_alpha(x)=g(x+alpha)-g(x)~forallalphainmathbbR
endequation*
and applying the Fourier transform (as well as a little manipulation).
The Denjoy-Perron integral is used instead of the Lebesgue integral to avoid the problem of a locally integrable function $f:mathbbRtomathbbC$ admitting a derivative $f'(x)~forall xinmathbbR$ without $f'$ being locally integrable.
The case for the Denjoy-Perron integral is proved in a similar fashion.
Check out the following paper by J. Reyna
http://www.ams.org/journals/proc/1990-109-01/S0002-9939-1990-1007485-4/S0002-9939-1990-1007485-4.pdf
answered Jun 20 '15 at 20:32
George Simpson
4,56432449
answered Jun 20 '15 at 20:32
George Simpson
4,56432449
answered Jun 20 '15 at 20:32
George Simpson
4,56432449
answered Jun 20 '15 at 20:32
George Simpson
4,56432449
4,56432449
add a comment |Â
add a comment |Â
up vote
8
down vote
The following theorem is a beautiful generalization of Frullani’s integral theorem.
Let $f(x)-f(infty)=sum_k=0^inftyfracu(k)(-x)^kk!$ and $g(x)-g(infty)=sum_k=0^inftyfracv(k)(-x)^kk!$
$Theorem1$:
Let f, and g be continuous function on $[0,infty),$ assume that $f(0)=g(0)$ and $f(infty)=g(infty)$. Then if $a,b>0$
$$lim_n to 0+I_nequiv lim_n to 0+ int_0^inftyx^n-1lbrace f(ax)-g(bx) rbrace dx=lbrace f(0)-f(infty)rbrace bigg lbrace log bigg(fracba bigg)+fracddsbigg(logbigg(fracv(s)u(s)bigg) bigg)_s=0 bigg rbrace$$
if $f(x)=g(x),$ this theorem reduces to the Frullani’s theorem
$$int_0^infty fracf(ax)-f(bx)xdx=lbrace f(0)-f(infty) rbrace log bigg(fracba bigg).$$
Let prove $Theorem1$, To do this we need to use Ramanujan's master theorem , Which lies in the fact that
$$int_0^infty x^n-1sum_k=0^infty frac phi(k)(-x)^kk!dx= Gamma(n)phi(-n).$$
Applying the Master Theorem with $0<n<1,$ we find
$$I_n=int_0^infty x^n-1( f(ax)-g(bx))dx=int_0^infty x^n-1( lbrace f(ax)-f(infty) rbrace-lbrace g(bx)-g(infty) rbrace) dx$$
$$=Gamma(n)lbrace a^-nu(-n)-b^-nv(-n) rbrace$$
$$=Gamma(n+1) bigg lbrace fraca^-nu(-n)-b^-nv(-n)n bigg rbrace
$$
Letting $n$ tend to $0$, using L'Hospital's Rule and fact that $u(0)=v(0)=f(0)-f(infty).$ we deduce that
$$lim_n to inftyI_n=lim_n to infty bigg lbrace fracb^nv(n)-a^nu(n)n bigg rbrace$$
$$=lim_n to infty lbrace b^nv(n) log b+ b^nv'(n)-a^nu(n)log a-a^nu'(n)rbrace$$
$$= lbrace f(0)-f(infty) rbrace log bigg(fracba bigg)+v'(0)-u'(0)$$
$$=lbrace f(0)-f(infty) rbrace bigg lbrace log bigg(fracba bigg)+fracddsbigg(logbigg(fracv(s)u(s)bigg) bigg)_s=0 bigg rbrace$$
answered Mar 16 '16 at 17:37
vito
945815
2
This is incredible! May I ask is there a source for this? I'd like to know about some related analysis.
– Lee David Chung Lin
Mar 9 at 11:29
add a comment |Â
up vote
8
down vote
The following theorem is a beautiful generalization of Frullani’s integral theorem.
Let $f(x)-f(infty)=sum_k=0^inftyfracu(k)(-x)^kk!$ and $g(x)-g(infty)=sum_k=0^inftyfracv(k)(-x)^kk!$
$Theorem1$:
Let f, and g be continuous function on $[0,infty),$ assume that $f(0)=g(0)$ and $f(infty)=g(infty)$. Then if $a,b>0$
$$lim_n to 0+I_nequiv lim_n to 0+ int_0^inftyx^n-1lbrace f(ax)-g(bx) rbrace dx=lbrace f(0)-f(infty)rbrace bigg lbrace log bigg(fracba bigg)+fracddsbigg(logbigg(fracv(s)u(s)bigg) bigg)_s=0 bigg rbrace$$
if $f(x)=g(x),$ this theorem reduces to the Frullani’s theorem
$$int_0^infty fracf(ax)-f(bx)xdx=lbrace f(0)-f(infty) rbrace log bigg(fracba bigg).$$
Let prove $Theorem1$, To do this we need to use Ramanujan's master theorem , Which lies in the fact that
$$int_0^infty x^n-1sum_k=0^infty frac phi(k)(-x)^kk!dx= Gamma(n)phi(-n).$$
Applying the Master Theorem with $0<n<1,$ we find
$$I_n=int_0^infty x^n-1( f(ax)-g(bx))dx=int_0^infty x^n-1( lbrace f(ax)-f(infty) rbrace-lbrace g(bx)-g(infty) rbrace) dx$$
$$=Gamma(n)lbrace a^-nu(-n)-b^-nv(-n) rbrace$$
$$=Gamma(n+1) bigg lbrace fraca^-nu(-n)-b^-nv(-n)n bigg rbrace
$$
Letting $n$ tend to $0$, using L'Hospital's Rule and fact that $u(0)=v(0)=f(0)-f(infty).$ we deduce that
$$lim_n to inftyI_n=lim_n to infty bigg lbrace fracb^nv(n)-a^nu(n)n bigg rbrace$$
$$=lim_n to infty lbrace b^nv(n) log b+ b^nv'(n)-a^nu(n)log a-a^nu'(n)rbrace$$
$$= lbrace f(0)-f(infty) rbrace log bigg(fracba bigg)+v'(0)-u'(0)$$
$$=lbrace f(0)-f(infty) rbrace bigg lbrace log bigg(fracba bigg)+fracddsbigg(logbigg(fracv(s)u(s)bigg) bigg)_s=0 bigg rbrace$$
answered Mar 16 '16 at 17:37
vito
945815
2
This is incredible! May I ask is there a source for this? I'd like to know about some related analysis.
– Lee David Chung Lin
Mar 9 at 11:29
add a comment |Â
up vote
8
down vote
up vote
8
down vote
The following theorem is a beautiful generalization of Frullani’s integral theorem.
Let $f(x)-f(infty)=sum_k=0^inftyfracu(k)(-x)^kk!$ and $g(x)-g(infty)=sum_k=0^inftyfracv(k)(-x)^kk!$
$Theorem1$:
Let f, and g be continuous function on $[0,infty),$ assume that $f(0)=g(0)$ and $f(infty)=g(infty)$. Then if $a,b>0$
$$lim_n to 0+I_nequiv lim_n to 0+ int_0^inftyx^n-1lbrace f(ax)-g(bx) rbrace dx=lbrace f(0)-f(infty)rbrace bigg lbrace log bigg(fracba bigg)+fracddsbigg(logbigg(fracv(s)u(s)bigg) bigg)_s=0 bigg rbrace$$
if $f(x)=g(x),$ this theorem reduces to the Frullani’s theorem
$$int_0^infty fracf(ax)-f(bx)xdx=lbrace f(0)-f(infty) rbrace log bigg(fracba bigg).$$
Let prove $Theorem1$, To do this we need to use Ramanujan's master theorem , Which lies in the fact that
$$int_0^infty x^n-1sum_k=0^infty frac phi(k)(-x)^kk!dx= Gamma(n)phi(-n).$$
Applying the Master Theorem with $0<n<1,$ we find
$$I_n=int_0^infty x^n-1( f(ax)-g(bx))dx=int_0^infty x^n-1( lbrace f(ax)-f(infty) rbrace-lbrace g(bx)-g(infty) rbrace) dx$$
$$=Gamma(n)lbrace a^-nu(-n)-b^-nv(-n) rbrace$$
$$=Gamma(n+1) bigg lbrace fraca^-nu(-n)-b^-nv(-n)n bigg rbrace
$$
Letting $n$ tend to $0$, using L'Hospital's Rule and fact that $u(0)=v(0)=f(0)-f(infty).$ we deduce that
$$lim_n to inftyI_n=lim_n to infty bigg lbrace fracb^nv(n)-a^nu(n)n bigg rbrace$$
$$=lim_n to infty lbrace b^nv(n) log b+ b^nv'(n)-a^nu(n)log a-a^nu'(n)rbrace$$
$$= lbrace f(0)-f(infty) rbrace log bigg(fracba bigg)+v'(0)-u'(0)$$
$$=lbrace f(0)-f(infty) rbrace bigg lbrace log bigg(fracba bigg)+fracddsbigg(logbigg(fracv(s)u(s)bigg) bigg)_s=0 bigg rbrace$$
answered Mar 16 '16 at 17:37
vito
945815
The following theorem is a beautiful generalization of Frullani’s integral theorem.
Let $f(x)-f(infty)=sum_k=0^inftyfracu(k)(-x)^kk!$ and $g(x)-g(infty)=sum_k=0^inftyfracv(k)(-x)^kk!$
$Theorem1$:
Let f, and g be continuous function on $[0,infty),$ assume that $f(0)=g(0)$ and $f(infty)=g(infty)$. Then if $a,b>0$
$$lim_n to 0+I_nequiv lim_n to 0+ int_0^inftyx^n-1lbrace f(ax)-g(bx) rbrace dx=lbrace f(0)-f(infty)rbrace bigg lbrace log bigg(fracba bigg)+fracddsbigg(logbigg(fracv(s)u(s)bigg) bigg)_s=0 bigg rbrace$$
if $f(x)=g(x),$ this theorem reduces to the Frullani’s theorem
$$int_0^infty fracf(ax)-f(bx)xdx=lbrace f(0)-f(infty) rbrace log bigg(fracba bigg).$$
Let prove $Theorem1$, To do this we need to use Ramanujan's master theorem , Which lies in the fact that
$$int_0^infty x^n-1sum_k=0^infty frac phi(k)(-x)^kk!dx= Gamma(n)phi(-n).$$
Applying the Master Theorem with $0<n<1,$ we find
$$I_n=int_0^infty x^n-1( f(ax)-g(bx))dx=int_0^infty x^n-1( lbrace f(ax)-f(infty) rbrace-lbrace g(bx)-g(infty) rbrace) dx$$
$$=Gamma(n)lbrace a^-nu(-n)-b^-nv(-n) rbrace$$
$$=Gamma(n+1) bigg lbrace fraca^-nu(-n)-b^-nv(-n)n bigg rbrace
$$
Letting $n$ tend to $0$, using L'Hospital's Rule and fact that $u(0)=v(0)=f(0)-f(infty).$ we deduce that
$$lim_n to inftyI_n=lim_n to infty bigg lbrace fracb^nv(n)-a^nu(n)n bigg rbrace$$
$$=lim_n to infty lbrace b^nv(n) log b+ b^nv'(n)-a^nu(n)log a-a^nu'(n)rbrace$$
$$= lbrace f(0)-f(infty) rbrace log bigg(fracba bigg)+v'(0)-u'(0)$$
$$=lbrace f(0)-f(infty) rbrace bigg lbrace log bigg(fracba bigg)+fracddsbigg(logbigg(fracv(s)u(s)bigg) bigg)_s=0 bigg rbrace$$
answered Mar 16 '16 at 17:37
vito
945815
answered Mar 16 '16 at 17:37
vito
945815
answered Mar 16 '16 at 17:37
vito
945815
answered Mar 16 '16 at 17:37
vito
945815
945815
2
This is incredible! May I ask is there a source for this? I'd like to know about some related analysis.
– Lee David Chung Lin
Mar 9 at 11:29
add a comment |Â
2
This is incredible! May I ask is there a source for this? I'd like to know about some related analysis.
– Lee David Chung Lin
Mar 9 at 11:29
2
2
This is incredible! May I ask is there a source for this? I'd like to know about some related analysis.
– Lee David Chung Lin
Mar 9 at 11:29
This is incredible! May I ask is there a source for this? I'd like to know about some related analysis.
– Lee David Chung Lin
Mar 9 at 11:29
add a comment |Â
up vote
2
down vote
On the assumption that $f$ is differentiable and $a,b>0$, an application of Fubini's theorem gives the result.
$$int_(0,infty) fracf(ax)-f(bx)x dx$$
$$=int_(0,infty) int_[bx,ax] f'(y) frac1x dy dx$$
Let $0<bx leq y leq ax$.
$$=int_(0,infty) int_frac1a y^frac1b y f'(y) frac1x dx dy$$
$$=int_(0,infty) f'(y) ln (fracab) dy$$
$$=(f(0)-f(infty)) ln fracba$$
answered Jul 23 '17 at 20:25
Ahmed S. Attaalla
13.8k11644
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On the assumption that $f$ is differentiable and $a,b>0$, an application of Fubini's theorem gives the result.
$$int_(0,infty) fracf(ax)-f(bx)x dx$$
$$=int_(0,infty) int_[bx,ax] f'(y) frac1x dy dx$$
Let $0<bx leq y leq ax$.
$$=int_(0,infty) int_frac1a y^frac1b y f'(y) frac1x dx dy$$
$$=int_(0,infty) f'(y) ln (fracab) dy$$
$$=(f(0)-f(infty)) ln fracba$$
answered Jul 23 '17 at 20:25
Ahmed S. Attaalla
13.8k11644
add a comment |Â
up vote
2
down vote
up vote
2
down vote
On the assumption that $f$ is differentiable and $a,b>0$, an application of Fubini's theorem gives the result.
$$int_(0,infty) fracf(ax)-f(bx)x dx$$
$$=int_(0,infty) int_[bx,ax] f'(y) frac1x dy dx$$
Let $0<bx leq y leq ax$.
$$=int_(0,infty) int_frac1a y^frac1b y f'(y) frac1x dx dy$$
$$=int_(0,infty) f'(y) ln (fracab) dy$$
$$=(f(0)-f(infty)) ln fracba$$
answered Jul 23 '17 at 20:25
Ahmed S. Attaalla
13.8k11644
On the assumption that $f$ is differentiable and $a,b>0$, an application of Fubini's theorem gives the result.
$$int_(0,infty) fracf(ax)-f(bx)x dx$$
$$=int_(0,infty) int_[bx,ax] f'(y) frac1x dy dx$$
Let $0<bx leq y leq ax$.
$$=int_(0,infty) int_frac1a y^frac1b y f'(y) frac1x dx dy$$
$$=int_(0,infty) f'(y) ln (fracab) dy$$
$$=(f(0)-f(infty)) ln fracba$$
answered Jul 23 '17 at 20:25
Ahmed S. Attaalla
13.8k11644
edited Jul 23 '17 at 20:34
answered Jul 23 '17 at 20:25
Ahmed S. Attaalla
13.8k11644
answered Jul 23 '17 at 20:25
Ahmed S. Attaalla
13.8k11644
answered Jul 23 '17 at 20:25
Ahmed S. Attaalla
13.8k11644
13.8k11644
add a comment |Â
add a comment |Â
up vote
1
down vote
The following is just a speeded-up version of the answer by Davide Giraudo.
Let $f(x)$ be a real-valued function defined for $xgeq 0$.
Suppose that $f(x)$ is Riemann integrable on every bounded interval
of nonnegative real numbers,
that $f(x)$ is continuous at $x=0$,
and that the limit $f(infty):=lim_xtoinfty f(x)$ exists
(as a finite quantity).
If $a>0$ and $b>0$, then the integral
beginequation*
int_,0^,infty fracf(ax)-f(bx)x dx tag1
endequation*
exists and is equal to $bigl(f(infty)-f(0)bigr),ln(a/b)$.
The assertion that the integral $(1)$ exists means
that the following limit exists,
beginequation*
lim_lto0,,htoinfty int_,l^,h fracf(ax)-f(bx)x dx
endequation*
where $l$ approaches $0$ independently of $h$ approaching $infty$,
and that this limit is the integral $(1)$.
Proof.$,$ Assume that $ageq b,$; this does not lose us any generality.
Let $0<l<h,$.
The change of variables $ax=by$ shows that
beginequation*
int_,l/a^,h/afracf(ax)xdx
~=~ int_,l/b^,h/bfracf(by)ydy~,
endequation*
so we have
beginalign*
int_,l/a^,h/afracf(ax)-f(bx)xdx
&~=~ int_,l/b^,h/bfracf(bx)xdx
~-~ int_,l/a^,h/afracf(bx)xdx \
&~=~ int_,h/a^,h/bfracf(bx)xdx
~-~ int_,l/a^,l/bfracf(bx)xdx~;
endalign*
we write the difference in the second line as $I(h) - I(l)$.
Let $varepsilon>0$.
There exists $l_varepsilon>0$ such that
$f(0)-varepsilonleq f(bx)leq f(0)+varepsilon$
for $0leq xleq l_varepsilon/b$.
Since by assumption $l/aleq l/b$ for every $l>0$, we obtain the estimate
beginequation*
int_,l/a^,l/bfracf(0)-varepsilonxdx
~leq~ I(l)
~leq~ int_,l/a^,l/bfracf(0)+varepsilonxdx
qquadqquad textfor $0<lleq l_varepsilon$~,
endequation*
that is,
beginequation*
bigl(f(0)-varepsilonbigr),lnfracab
~leq~ I(l)
~leq~ bigl(f(0)+varepsilonbigr),lnfracab
qquadqquad textfor $0<lleq l_varepsilon$~.
endequation*
In other words, $I(l)$ converges to $f(0),ln(a/b)$ as $l$ approaches $0$.
In the same way we see that $I(h)$ converges to $f(infty)$
as $h$ approaches $infty$.$,$ Done.
answered Jan 3 '17 at 14:57
chizhek
2,587622
add a comment |Â
up vote
1
down vote
The following is just a speeded-up version of the answer by Davide Giraudo.
Let $f(x)$ be a real-valued function defined for $xgeq 0$.
Suppose that $f(x)$ is Riemann integrable on every bounded interval
of nonnegative real numbers,
that $f(x)$ is continuous at $x=0$,
and that the limit $f(infty):=lim_xtoinfty f(x)$ exists
(as a finite quantity).
If $a>0$ and $b>0$, then the integral
beginequation*
int_,0^,infty fracf(ax)-f(bx)x dx tag1
endequation*
exists and is equal to $bigl(f(infty)-f(0)bigr),ln(a/b)$.
The assertion that the integral $(1)$ exists means
that the following limit exists,
beginequation*
lim_lto0,,htoinfty int_,l^,h fracf(ax)-f(bx)x dx
endequation*
where $l$ approaches $0$ independently of $h$ approaching $infty$,
and that this limit is the integral $(1)$.
Proof.$,$ Assume that $ageq b,$; this does not lose us any generality.
Let $0<l<h,$.
The change of variables $ax=by$ shows that
beginequation*
int_,l/a^,h/afracf(ax)xdx
~=~ int_,l/b^,h/bfracf(by)ydy~,
endequation*
so we have
beginalign*
int_,l/a^,h/afracf(ax)-f(bx)xdx
&~=~ int_,l/b^,h/bfracf(bx)xdx
~-~ int_,l/a^,h/afracf(bx)xdx \
&~=~ int_,h/a^,h/bfracf(bx)xdx
~-~ int_,l/a^,l/bfracf(bx)xdx~;
endalign*
we write the difference in the second line as $I(h) - I(l)$.
Let $varepsilon>0$.
There exists $l_varepsilon>0$ such that
$f(0)-varepsilonleq f(bx)leq f(0)+varepsilon$
for $0leq xleq l_varepsilon/b$.
Since by assumption $l/aleq l/b$ for every $l>0$, we obtain the estimate
beginequation*
int_,l/a^,l/bfracf(0)-varepsilonxdx
~leq~ I(l)
~leq~ int_,l/a^,l/bfracf(0)+varepsilonxdx
qquadqquad textfor $0<lleq l_varepsilon$~,
endequation*
that is,
beginequation*
bigl(f(0)-varepsilonbigr),lnfracab
~leq~ I(l)
~leq~ bigl(f(0)+varepsilonbigr),lnfracab
qquadqquad textfor $0<lleq l_varepsilon$~.
endequation*
In other words, $I(l)$ converges to $f(0),ln(a/b)$ as $l$ approaches $0$.
In the same way we see that $I(h)$ converges to $f(infty)$
as $h$ approaches $infty$.$,$ Done.
answered Jan 3 '17 at 14:57
chizhek
2,587622
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The following is just a speeded-up version of the answer by Davide Giraudo.
Let $f(x)$ be a real-valued function defined for $xgeq 0$.
Suppose that $f(x)$ is Riemann integrable on every bounded interval
of nonnegative real numbers,
that $f(x)$ is continuous at $x=0$,
and that the limit $f(infty):=lim_xtoinfty f(x)$ exists
(as a finite quantity).
If $a>0$ and $b>0$, then the integral
beginequation*
int_,0^,infty fracf(ax)-f(bx)x dx tag1
endequation*
exists and is equal to $bigl(f(infty)-f(0)bigr),ln(a/b)$.
The assertion that the integral $(1)$ exists means
that the following limit exists,
beginequation*
lim_lto0,,htoinfty int_,l^,h fracf(ax)-f(bx)x dx
endequation*
where $l$ approaches $0$ independently of $h$ approaching $infty$,
and that this limit is the integral $(1)$.
Proof.$,$ Assume that $ageq b,$; this does not lose us any generality.
Let $0<l<h,$.
The change of variables $ax=by$ shows that
beginequation*
int_,l/a^,h/afracf(ax)xdx
~=~ int_,l/b^,h/bfracf(by)ydy~,
endequation*
so we have
beginalign*
int_,l/a^,h/afracf(ax)-f(bx)xdx
&~=~ int_,l/b^,h/bfracf(bx)xdx
~-~ int_,l/a^,h/afracf(bx)xdx \
&~=~ int_,h/a^,h/bfracf(bx)xdx
~-~ int_,l/a^,l/bfracf(bx)xdx~;
endalign*
we write the difference in the second line as $I(h) - I(l)$.
Let $varepsilon>0$.
There exists $l_varepsilon>0$ such that
$f(0)-varepsilonleq f(bx)leq f(0)+varepsilon$
for $0leq xleq l_varepsilon/b$.
Since by assumption $l/aleq l/b$ for every $l>0$, we obtain the estimate
beginequation*
int_,l/a^,l/bfracf(0)-varepsilonxdx
~leq~ I(l)
~leq~ int_,l/a^,l/bfracf(0)+varepsilonxdx
qquadqquad textfor $0<lleq l_varepsilon$~,
endequation*
that is,
beginequation*
bigl(f(0)-varepsilonbigr),lnfracab
~leq~ I(l)
~leq~ bigl(f(0)+varepsilonbigr),lnfracab
qquadqquad textfor $0<lleq l_varepsilon$~.
endequation*
In other words, $I(l)$ converges to $f(0),ln(a/b)$ as $l$ approaches $0$.
In the same way we see that $I(h)$ converges to $f(infty)$
as $h$ approaches $infty$.$,$ Done.
answered Jan 3 '17 at 14:57
chizhek
2,587622
The following is just a speeded-up version of the answer by Davide Giraudo.
Let $f(x)$ be a real-valued function defined for $xgeq 0$.
Suppose that $f(x)$ is Riemann integrable on every bounded interval
of nonnegative real numbers,
that $f(x)$ is continuous at $x=0$,
and that the limit $f(infty):=lim_xtoinfty f(x)$ exists
(as a finite quantity).
If $a>0$ and $b>0$, then the integral
beginequation*
int_,0^,infty fracf(ax)-f(bx)x dx tag1
endequation*
exists and is equal to $bigl(f(infty)-f(0)bigr),ln(a/b)$.
The assertion that the integral $(1)$ exists means
that the following limit exists,
beginequation*
lim_lto0,,htoinfty int_,l^,h fracf(ax)-f(bx)x dx
endequation*
where $l$ approaches $0$ independently of $h$ approaching $infty$,
and that this limit is the integral $(1)$.
Proof.$,$ Assume that $ageq b,$; this does not lose us any generality.
Let $0<l<h,$.
The change of variables $ax=by$ shows that
beginequation*
int_,l/a^,h/afracf(ax)xdx
~=~ int_,l/b^,h/bfracf(by)ydy~,
endequation*
so we have
beginalign*
int_,l/a^,h/afracf(ax)-f(bx)xdx
&~=~ int_,l/b^,h/bfracf(bx)xdx
~-~ int_,l/a^,h/afracf(bx)xdx \
&~=~ int_,h/a^,h/bfracf(bx)xdx
~-~ int_,l/a^,l/bfracf(bx)xdx~;
endalign*
we write the difference in the second line as $I(h) - I(l)$.
Let $varepsilon>0$.
There exists $l_varepsilon>0$ such that
$f(0)-varepsilonleq f(bx)leq f(0)+varepsilon$
for $0leq xleq l_varepsilon/b$.
Since by assumption $l/aleq l/b$ for every $l>0$, we obtain the estimate
beginequation*
int_,l/a^,l/bfracf(0)-varepsilonxdx
~leq~ I(l)
~leq~ int_,l/a^,l/bfracf(0)+varepsilonxdx
qquadqquad textfor $0<lleq l_varepsilon$~,
endequation*
that is,
beginequation*
bigl(f(0)-varepsilonbigr),lnfracab
~leq~ I(l)
~leq~ bigl(f(0)+varepsilonbigr),lnfracab
qquadqquad textfor $0<lleq l_varepsilon$~.
endequation*
In other words, $I(l)$ converges to $f(0),ln(a/b)$ as $l$ approaches $0$.
In the same way we see that $I(h)$ converges to $f(infty)$
as $h$ approaches $infty$.$,$ Done.
answered Jan 3 '17 at 14:57
chizhek
2,587622
answered Jan 3 '17 at 14:57
chizhek
2,587622
answered Jan 3 '17 at 14:57
chizhek
2,587622
answered Jan 3 '17 at 14:57
chizhek
2,587622
2,587622
add a comment |Â
add a comment |Â
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The answers below are all good. There is an AMS article (1990) at: ams.org/journals/proc/1990-109-01/S0002-9939-1990-1007485-4 and it's free//available!
– rrogers
Apr 21 at 14:27